Limit with little-o It's a while I'm trying to verify this limit:
$$\lim_{x\rightarrow 0^+} \frac{\log(x+\sqrt{1+x^2})-\sin x}{(2^{x^2}-1)(\sin (e^{x^2}-1))} = 0,$$
this is my attempt:
Since $x \rightarrow 0$ I can do the substitution:
$$\sqrt{1+x^2}=1+o(1)$$
$$\sin{x}=x(1+o(1))$$
and I obtain:
$$\frac{\log(x+1+o(1))-x(1+o(1))}{(2^{x^2}-1)(\sin (e^{x^2}-1))} = 0$$
Now I do these substitutions for $y \rightarrow 0$
$$\log (y+1)= y(1+o(1))$$
$$a^y-1 = y(\log a + o(1))$$
$$\sin y = y(1 + o(1))$$
and I obtain:
$$\frac{(x+o(1))(1+o(1))-x(1+o(1))}{x^2(\log 2+o(1))x^2(1+o(1))}=\frac{x+o(1)-x+o(x)}{(x^2 \log 2 + o(x^2))(x^2+o(x^2))}=\frac{o(1)}{x^4 \log 2 + o(x^4)}.$$
But the limit with $x \rightarrow 0$ of this makes infinite, not $0$ as it should.
Where is my error? 
Thank you.
 A: You have
\begin{align}
\sqrt{1+x^2}&=1+\frac{x^2}2-\frac{x^4}8+o(x^6)\\ \ \\
\log(1+x)&=x-\frac{x^2}2+\frac{x^3}3+o(x^4)\\ \ \\
\sin x&=x-x^3/6+o(x^5)\\ \ \\
2^x&=1+x\log2+o(x^2)\\ \ \\
e^x&=1+x+o(x^2)
\end{align}
Then
\begin{align}
\frac{\log(x+\sqrt{1+x^2})-\sin x}{(2^{x^2}-1)(\sin (e^{x^2}-1))}
&=\frac{\log(1+x+x^2/2-x^4/8+o(x^6))-(x-x^3/6+o(x^5))}{(x^2\log2+o(x^4))(\sin(x^2+o(x^4)))}\\ \ \\
&=\frac{\log(x+\sqrt{1+x^2})-\sin x}{(x^2\log2+o(x^4))(x^2+o(x^4))}\\ \ \\
&=\frac{\log(x+\sqrt{1+x^2})-\sin x}{o(x^4)}.
\end{align}
So the denominator is clearly $o(x^4)$. Let us work on the numerator: using Wolphram Alpha,
\begin{align}
\log(x+\sqrt{1+x^2})&=\log(1+x+\frac{x^2}2-\frac{x^4}8+o(x^6))\\ \ \\
&=x+\frac{x^2}2-\frac{x^4}8-\frac{(x+\frac{x^2}2-\frac{x^4}8)^2}2
+\frac{(x+\frac{x^2}2-\frac{x^4}8)^3}3+o(x^6)\\ \ \\
&=x-\frac{x^3}6+o(x^5).
\end{align}
Then
$$
\log(x+\sqrt{1+x^2})-\sin x=x-\frac{x^3}6+o(x^5)-\left(x-\frac{x^3}6+o(x^5)\right)=o(x^5).
$$
Thus
$$
\frac{\log(x+\sqrt{1+x^2})-\sin x}{(2^{x^2}-1)(\sin (e^{x^2}-1))}=\frac{o(x^5)}{o(x^4)}=o(x).
$$
